Conservation of Momentum of a box of mass

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SUMMARY

The discussion revolves around the conservation of momentum in a collision between two boxes, specifically Box A (1.20 kg) and Box B (2.56 kg). Box A moves at 2.52 m/s before colliding with Box B and bounces back at 0.665 m/s. The momentum of Box A before the collision is calculated as 3.02 kg*m/s to the right, while after the collision, it is 0.798 kg*m/s to the left. The correct calculation for the momentum of Box B after the collision is determined to be 3.818 kg*m/s to the right.

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blue_lilly
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Homework Statement


Box A of mass 1.20 kg is sliding to the right across a frictionless table at a speed of 2.52 m/s. Box A collides with Box B which has a mass of 2.56 kg, and Box A bounces straight back to the left with a speed of 0.665 m/s.
A) What is the momentum of Box A before the collision?
3.02 kg*m/s To the Right Correct Answer​
B) What is the momentum of Box A after the collision?
0.798 kg*m/s Correct: To the Left Correct Answer​
C) What is the momentum of Box B after the collision? This is the one I need help with!

Homework Equations


Momentum(P)=mass(kg)*Velocity(kg*m/s)
Momentum(P-before)=Momentum(P-after)

The Attempt at a Solution


A) P=m*V=(1.20 kg)(2.52 m/s)=3.02 kg*m/s To the right CORRECT
B) P=m*V=(1.20 kg)(.665 m/s)=.798 kg*m/s To the Left CORRECT
This is the part I am having trouble with
C) Momentum(before)=Momentum(after); P=m*V
PBeforeA+PBeforeB=PAfterA+PAfterB
[(1.20 kg)(2.52 m/s)]+[(2.56 kg)(0 m/s)] = [(1.20 kg)(.665 m/s)]+[]
[3.02 kg*m/s]+[0 kg*m/s] = [.798 kg*m/s]+[PAfterB]
[(3.02 kg*m/s)/(.798 kg*m/s)] = PAfterB
PAfterB = [(3.02 kg*m/s)/(.798 kg*m/s)]
PAfterB = 3.78 kg*m/s : To the Right

The answer is incorrect. I am assuming the [To the right] portion is correct because BoxA and BoxB collided; this means that they are asserting equal and opposite force on each other and because BoxA went to the Left, BoxB must go to the Right.

This leaves the incorrect portion to be the Momentum of BoxB after the collision. I double checked my math, a numerous amount of time, so I don't know where I am going wrong.

Help is appreciated!
 
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blue_lilly said:
[(1.20 kg)(2.52 m/s)]+[(2.56 kg)(0 m/s)] = [(1.20 kg)(.665 m/s)]+[]
Decide which direction, left or right, is positive. Say it's right. If I tell you box A has momentum +0.8kgm/s at some point, which way is it moving?
 
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haruspex said:
Decide which direction, left or right, is positive. Say it's right. If I tell you box A has momentum +0.8kgm/s at some point, which way is it moving?

So if I say the right is positive, then anytime a box is moving to the right-there would be a + sign and if it was moving to the left, a -sign.

[+3.02 kg*m/s]+[0 kg*m/s] = [-.798 kg*m/s]+[+PAfterB]
[(+3.02 kg*m/s)-(-.798 kg*m/s)] = +PAfterB
+PAfterB = [(+3.02 kg*m/s)-(-.798 kg*m/s)]
PAfterB = 3.818 kg*m/s To The Right

Is this correct?
 
blue_lilly said:
PAfterB = 3.818 kg*m/s To The Right
Looks right.
(You also did some very strange things later on in the OP, like dividing instead of subtracting. But you got it right this time.)
 
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haruspex said:
Looks right.
(You also did some very strange things later on in the OP, like dividing instead of subtracting. But you got it right this time.)

Thank you so much, that makes more sense!

I caught the dividing error after I posted the question to the forum! C:
 

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